# Proving hyperbolic identities

I am supposed to prove that $\tanh\ln x = \dfrac {x^2 - 1} {x^2+1}$. As far as I can tell this is not true.

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We try to do all the details. By definition, $$\sinh(u)=\frac{e^u-e^{-u}}{2} \text{ and } \cosh(u)=\frac{e^u+e^{-u}}{2}.$$

Set $u=\ln x$, and use the fact that $e^{\ln x}=x$ and $e^{-\ln x}=\frac{1}{e^{\ln x}}=\frac{1}{x}$. We get $$\sinh(\ln x)=\frac{x-\frac{1}{x}}{2} \text{ and } \cosh(\ln x)=\frac{x+\frac{1}{x}}{2}.$$ These respectively simplify to $\frac{x^2-1}{2x}$ and $\frac{x^2+1}{2x}$. Here are the details for $\sinh(\ln x)$.

We want to simplify $\frac{x-\frac{1}{x}}{2}$. Look first at the numerator $x-\frac{1}{x}$. The second part has denominator $x$. We want the first part also to have denominator $x$, so that both parts will have a common denominator.

We have $x=\frac{x}{1}$. Multiply top and bottom by $x$. So $x=\frac{x}{1}=\frac{x^2}{x}$.

We conclude that $x-\frac{1}{x}=\frac{x^2}{x}-\frac{1}{x}=\frac{x^2-1}{x}$. Now back to $\sinh(\ln x)$. We have $$\sinh(\ln x)=\frac{x-\frac{1}{x}}{2}=\frac{\frac{x^2-1}{x}}{2}=\frac{x^2-1}{2x}.$$ A very similar calculation takes care of $\cosh(\ln x)$.

Finally, use $\tanh(u)=\frac{\sinh(u)}{\cosh(u)}$ (this is just the definition of $\tanh(u)$). We get $$\tanh(\ln x)=\frac{\sinh(\ln x)}{\cosh(\ln x)}=\frac{\frac{x^2-1}{2x}}{\frac{x^2+1}{2x}}.$$ Do the division. The $2x$ parts "cancel". Alternately, multiply top and bottom by $2x$. We get $$\tanh(\ln x)=\frac{x^2-1}{x^2+1}.$$

Comment: The derivation looks (and is) long. But that's because the steps that involve basic algebra were done in extreme detail. With some experience, one can do these steps in one's head, and the calculation can be reduced to a couple of lines.

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I don't get it, so $e^{lnx}$ is like $e^{loge}$ so I am raising e to loge? And that gives me e? –  Jordan Oct 6 '11 at 21:15
No. $x=e^{\ln(x)} =e^{\log_e(x)}$ –  Henry Oct 6 '11 at 21:23
@Jordan: No, $e^{\ln x}=x$. It is basically $\exp(\log_e x)$, but that's not the same as $\exp(\log e)$. –  anon Oct 6 '11 at 21:23
I don't understand. –  Jordan Oct 6 '11 at 21:24
You absolutely need to know the basic facts $e^{\ln x}=x$ and $\ln(e^x)=x$. The $\ln(x)$ function and the $e^x$ function are inverses of each other. You will need to use these facts over and over, both for "theory" and for practical applications. You need in fact much more. The name $\ln x$ is another name for $\log_e(x)$. So $e^{\ln x}=e^{\log_e(x)}=x$. Is that what you were asking? I could not tell for sure. –  André Nicolas Oct 6 '11 at 21:28
You have to use the fact that $\tanh t=\frac{\sinh t}{\cosh t}$ and that $\sinh t=\frac{e^t-e^{-t}}2$ and $\cosh t=\frac{e^t+e^{-t}}2$. Can you see how to proceed?
Oh well... ${}{}{}$ –  Mariano Suárez-Alvarez Oct 6 '11 at 21:12